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The stumbled upon following problem from Sheldon Ross's book:

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue and 18 green balls. What is probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn?

In this pdf, the solution is given as $p=\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.

But I guess it should be $\frac{\binom{12}{3}\times\binom{16+18}{4}+\binom{16}{3}\times\binom{12+18}{4}-\color{red}{2\times}\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}}{\binom{12+16+18}{7}}$.

Am I right? (The book has not given answer.)

RajS
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  • The answer in the pdf is correct. Remember that $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$. If you subtract twice, you would be finding the probability that exactly one of the two events occurs. – N. F. Taussig Jun 15 '18 at 20:31
  • Why $2\times$? The first answer should be correct – Green.H Jun 15 '18 at 20:31
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    I think he's interpreting "either" as the symmetric diference between the sets instead of the union. – BelwarDissengulp Jun 15 '18 at 20:34
  • The problem says "either exactly 3 red balls or exactly 3 blue balls". Doesnt that means exclusive OR (exactly one of the two events)? Cant we add "but not both" at the end, that is "either exactly 3 red balls or exactly 3 blue balls (but not both)"? Doesnt this sound more correct / more inline with whats given? – RajS Jun 15 '18 at 20:36
  • It would be exclusive if you only drew 3, 4, or 5 balls. But since you're drawing 6 or more, the "or" has to be inclusive, unless Ross specifically states "but not both." The use of "exactly" is throwing you off. It limits the numbers of red and blue, but not the total. – Ted Shifrin Jun 15 '18 at 20:38
  • I didnt get "It would be exclusive if you only drew 3, 4, or 5 balls. But since you're drawing 6 or more, the "or" has to be inclusive". See N F Taussig also agrees, if I multiply by 2, I will be doing exclusive OR. That means exclusive OR is possible with given number of selection. Right? – RajS Jun 15 '18 at 20:41
  • If it was "3 red balls or 3 blue balls", it would have been union. But here it says "either exactly 3 red balls or exactly 3 blue balls". I feel "either" gives sense of exclusive OR? – RajS Jun 15 '18 at 20:44
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    Unfortunately, language is fluid and imprecise. If I could make everyone in the world agree that the phrase "Either the event A or B occurs" implies the symmetric difference $A\triangle B$, i.e. the exclusive or, I would. Unfortunately there are some that interpret "Either the event A or B" as the inclusive or, i.e. the union $A\cup B$. Some authors intend it the one way while other authors intend it the other way. As a result, I try to avoid the phrase without the use of additional qualifiers opting for "Either A or B but not both" if I mean the exclusive or. – JMoravitz Jun 15 '18 at 20:45
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    If the question is intended merely for practice, then practice by answering both interpretations and understand the difference between the meaning of the different interpretations and understand the difference between the solutions to the different interpretations. If the problem is intended for homework or a test, then ask for clarification from the professor which interpretation is intended if able. Otherwise, if unable to ask for clarification then state which interpretation you are using if it is potentially ambiguous and affects the outcome. – JMoravitz Jun 15 '18 at 20:46

2 Answers2

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I'd interpret the question as that exactly one of the two events (3 red or 3 blue) occurs but not both ("either or" strongly hints at that). This amounts to $P(3 \text{ red}) + P(3 \text{blue}) - 2P(3 \text{ red and } 3 \text{ blue })$ which is

$${{\binom{12}{3}\binom{16+18}{4}}\over{\binom{12+16+18}{7}}} + {{\binom{16}{3}\binom{12+18}{4}}\over {\binom{12+16+18}{7}}} - 2{{\binom{12}{3}\binom{16}{3}\binom{18}{1}} \over {\binom{12+16+18}{7}}}$$

In short, I concur.

Henno Brandsma
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    In North American mathematical English, "either...or" hints at nothing. It means the union ($\cup$) "...all of the points in A or B or both...all points that are in at least one of the sets...." If one specifies the exclusive 'or' then it's " either A or B, but not both." If you can quote a current, frequently used probability text where another convention is used, I'd be happy for the correction. (I have seen texts from the early 1940s that use $+$ and $\sum$ instead of $\cup$ when exclusive or is intended throughout a discussion.) – BruceET Jun 16 '18 at 00:54
  • For me "either-or" naturally means "either-or, but not both". Its so natural to me that I am unable to grasp how its not the same for English in other regions. But my intuition may not be that universally true and its just the fact that English in other regions have different meaning. Wikipedia points to the fact that Oxford dictionary interprets "either-or" as "exclusive OR". However, I also found math.stackexchage post, which says meaning is uncertain. – RajS Jun 16 '18 at 05:27
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    @BruceET See here for people defending my view of "either..or".@anir – Henno Brandsma Jun 16 '18 at 07:32
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Comment: The probability of exactly 3 red balls or exactly 3 blue balls (or both) from combinatorics is computed in R as 0.4359:

num = choose(12,3)*choose(34,4) + choose(16,3)*choose(30,4) - 
   choose(12,3)*choose(16,3)*choose(18,1)
ans = num/choose(46,7); ans
## 0.4359096

Simulation of 10 million draws of seven cards provides answers accurate to three places for both the inclusive and exclusive interpretations of or.

m = 10^7;  r = b = numeric(m)
urn = c(rep(1,12),rep(2,16),rep(3,18)) # 1=red, 2=blue, 3=green
for(i in 1:m) {
  draw=sample(urn,7)                   # sample without replacement
  r[i]=sum(draw==1);  b[i]=sum(draw==2) }
mean(r==3 | b==3)
## 0.4355743                           # union (inclusive 'or'): 3-place accuracy
mean(xor(r==3, b==3))
## 0.3941209                           # exclusive 'or'
mean(r==3 & b==3)
## 0.0414534                           # intersection

The same simulation provides marginal distributions of the the number of red balls drawn and the number of blue balls drawn, mostly accurate to three places. (Note that 0.000 represents a positive probability $< 0.0005.$ Specifically, ${12 \choose 7}/{46 \choose 7} = 1.4797 \times 10^{-5}$ and ${14 \choose 7}/{46 \choose 7} = 6.4120 \times 10^{-5}$.)

round(table(r)/m,3)
r
    0     1     2     3     4     5     6     7 
0.100 0.302 0.343 0.190 0.055 0.008 0.001 0.000 
round(table(b)/m,3)
b
    0     1     2     3     4     5     6     7 
0.038 0.178 0.319 0.287 0.138 0.035 0.004 0.000

In the joint distribution, values with $b + r > 7$ are, of course, exactly $0.$ Notice that the probability (approximately 0.041) of getting three balls of each color is a rounded version of 0.0414534 from above. (The exact value of this intersection from combinatorics is 0.04143135.)

round(table(r, b)/m, 3) 
   b
r       0     1     2     3     4     5     6     7
  0 0.001 0.006 0.019 0.032 0.028 0.012 0.003 0.000
  1 0.004 0.031 0.082 0.102 0.063 0.018 0.002 0.000
  2 0.011 0.060 0.121 0.106 0.040 0.005 0.000 0.000
  3 0.013 0.054 0.075 0.041 0.007 0.000 0.000 0.000
  4 0.008 0.023 0.020 0.005 0.000 0.000 0.000 0.000
  5 0.002 0.004 0.002 0.000 0.000 0.000 0.000 0.000
  6 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
  7 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
BruceET
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